Solution to a. Still working on b. Suppose $a,b \in (k^2, (k+1)^2)$ have a product of a perfect square. Write $a = m^2x$ and $b = n^2y$ where $x,y$ are squarefree. Remark then that $x=y$ if $ab$ is a perfect square, so $a = m^2x, b = n^2x$. WLOG $b > a$, so $n \ge m+1 \implies b-a \ge x(2m+1)$. However, we know $k^2 < m^2x, n^2x < k^2 + 2k + 1$ so $b-a \le 2k-1$. We also know $m > \frac{k}{\sqrt{x}}$. Thus $b-a \ge x \cdot \left ( \frac{2k}{\sqrt{x}} + 1 \right ) > 2k-1$, contradiction so $a,b$ do not exist.

Solution to b. We are done if there exists $k^n < a_1^{n-1} < a_2^{n-1} < ... < a_{n-1}^{n-1} < (k+1)^n$, for $a_i$ being positive integers, as then $k^n < a_1a_2...a_{n-1} < (k+1)^n$ and the product of $a_1^{n-1}, a_2^{n-1}, ... ,a_{n-1}^{n-1}, a_1a_2...a_{n-1}$ is a perfect $n$ th power. We claim that $a_1, a_2, ..., a_{n-1}$ as $a_1 = \lceil{k^{\frac{n}{n-1}} \rceil}$, and $a_{i+1}=a_i + 1$ for all $1 \le i \le n-2$, will work for sufficiently large $k$. The condition that needs to be fulfilled is $(a_1 + n-2)^{n-1} < (k+1)^n$ which is true if $(k^{\frac{n}{n-1}} + n-1)^{n-1} < (k+1)^n \Leftrightarrow n-1 < (k+1)^{\frac{n}{n-1}} - k^{\frac{n}{n-1}}$ which holds for sufficiently large $k$.

For the previous solution I think to avoid using this (or some ad hoc argument for this particular case) it's easiest to just take \[k^n<a_1^{n-1}<a_2^{n-1}<\cdots<a_{n-1}^{n-1}<a_n^{n-1}<(k+1)^n\]instead (again with the $a_i$ consecutive), and note that $a_1a_3a_4\cdots a_n$, $a_2a_3a_4\cdots a_n$ can't both be $n-1$th powers (or else $a_2/a_1=(a_1+1)/a_1$ would be). Here's another construction I found that's a bit messier but still instructive IMO.